MHB Find Vector Perpendicular to Plane

AI Thread Summary
To find a vector perpendicular to the plane defined by points P (1, 2, 3), Q (2, 3, 1), and R (3, 1, 2), the vector product of vectors PQ and PR is used. The vectors are calculated as PQ = (1, 1, -2) and PR = (2, -1, -1). The cross product PQ × PR is computed using the determinant of a matrix formed by these vectors. This method effectively yields a vector that is orthogonal to the plane formed by the three points. Understanding the vector product is essential for solving this problem.
brinlin
Messages
12
Reaction score
0
Find a vector that is perpendicular to the plane passing through the points P (1, 2, 3), Q (2, 3, 1), and R (3, 1, 2).
 
Mathematics news on Phys.org
vector product yields a vector perpendicular to two vectors ...

$\vec{PQ} \times \vec{PR}$
 
I'm sorry I don't really understand.
 
brinlin said:
I'm sorry I don't really understand.

you don't understand, or you don't know what a vector product is and how to calculate it?

$\vec{PQ} = (1,1,-2)$
$\vec{PR} = (2,-1,-1)$

$ \vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1& 1 & -2 \\ 2 &-1 &-1 \\ \end{vmatrix} $
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Thread 'Imaginary Pythagorus'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top